Description
Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence. We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.
Input
Input will consist of multiple input sets. Each set will consist of three integers, n, m, and d on a single line. The values of n, m and d will satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0 0 will indicate end of input and should not be processed.
For each input set, output a single line consisting of a comma-separated list of integers forming a degree danti-prime sequence (do not insert any spaces and do not split the output over multiple lines). In the case where more than one anti-prime sequence exists, print the lexicographically first one (i.e., output the one with the lowest first value; in case of a tie, the lowest second value, etc.). In the case where no anti-prime sequence exists, output No anti-prime sequence exists.
Sample Input
1 10 2 1 10 3 1 10 5 40 60 7 0 0 0
Sample Output
1,3,5,4,2,6,9,7,8,10 1,3,5,4,6,2,10,8,7,9 No anti-prime sequence exists. 40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54
分析：
本题的思路很直白，因为数据量并不是非常大，只是1000的范围，长度为10，那么和最大也只有10000以内，所以可以暴力搜索，对每一个节点都进行判断并且迭代搜索，明显是用DFS。注意先用筛法做个素数表辅助判断，直接进行素数判断太麻烦，会超时。这里看到别人的代码，说道DFS卡时，其实就是限制DFS的总搜索次数，模糊判断用来节省时间。当然这样答案并不准确，但是对于数据规模较小的时候，可以取一些极限值测试，本题4000次搜索就足够了。
代码：
1 // Problem#: 1002 2 // Submission#: 1792032 3 // The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License 4 // URI: http://creativecommons.org/licenses/by-nc-sa/3.0/ 5 // All Copyright reserved by Informatic Lab of Sun Yat-sen University 6 #include<iostream> 7 #include<cstring> 8 using namespace std; 9 10 #define MAX 10000 11 #define N 1010 12 13 bool isPrime[MAX]; 14 int prime[MAX]; 15 bool visit[N]; 16 int re[N]; 17 int n,m,d,t,sum; 18 bool flag; 19 20 void primeList(){ 21 memset(isPrime,true,sizeof(isPrime)); 22 for(int i = 2;i <= MAX;++i){ 23 if(isPrime[i]) prime[++prime[0]] = i; 24 for(int j = 1,k;j <= MAX && (k = i * prime[j]) <= MAX;++j) 25 { 26 isPrime[k] = false; 27 if(i % prime[j] == 0) break; 28 } 29 } 30 } 31 32 bool judge(int len){ 33 int sum; 34 if(len <= 1) return true; 35 for(int i = 0;i < len;++i){ 36 sum = 0; 37 if(i <= len - d){ 38 for(int j = i;j - i + 1<= d ;++j){ 39 sum += re[j]; 40 if(j-i+1 > 1 && isPrime[sum]) return false; 41 } 42 }else{ 43 for(int j = i;j < len;++j){ 44 sum += re[j]; 45 if(j-i+1 > 1 && isPrime[sum]) return false; 46 } 47 } 48 } 49 return true; 50 } 51 52 void dfs(int num){ 53 if(++t > 4000) return; 54 if(flag) return; 55 if(!judge(num)) return; 56 if(num== m - n + 1){ 57 flag = 1; 58 return; 59 } 60 for(int i = n;i <= m && !flag;++i){ 61 if(visit[i]) continue; 62 re[num] = i; 63 visit[i] = 1; 64 dfs(num+1); 65 visit[i] = 0; 66 } 67 } 68 69 int main(){ 70 primeList(); 71 while(cin>>n>>m>>d && n && m && d){ 72 flag = false; 73 t = 0; 74 memset(visit,0,sizeof(visit)); 75 dfs(0); 76 if(flag){ 77 cout << re[0]; 78 for( int i=1 ; i<m-n+1 ; ++i ) cout << "," << re[i]; 79 cout << endl; 80 }else cout << "No anti-prime sequence exists." << endl; 81 } 82 return 0; 83 }